An electric dipole of dipole moment →p consists of point charges +q and −q separated by a distance 2a apart. Deduce the expression for the electric field →E due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment →p. Hence show that in the limit X>>a, →E→2→p/(4πε₀x³). Given the electric field in the region →E=2x^i, find the net electric flux through the cube and the charge enclosed by it.<p></p>

(a) →p=2qa^i
→E+q=Kq(x−a)²^i
→E−q=−Kq(x+a)²^i
→E=(Kq(x−a)²−Kq(x+a)²) ^i
→E=4Kqax(x²−a²)²^i
→E=2K x →p(x²−a²)²
Using binomial expansion,
→E=2K→px³(1−x²a²+2x⁴a⁴−)
Neglecting higher terms,
→E=2→p4πε₀x³
(b)Since electric field is along x-axis, the only surfaces through which electric flux is non-zero are those parallel to the y-z plane.
At x=a, φ₁=∫→E.→dS
φ₁=−(2x)x=0 ×a²=0
φ₂=∫→E.→dS
φ₂=+(2x)x=a ×a²=2a³
Net flux is given by:
φ=2a³
By gauss law,
φ=qenclosed/ε₀
qenclosed=2a³ε₀

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